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Math Help - Solving some equations :)

  1. #1
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    Question Solving some equations :)

    Hello everyone, i have a few questions

    a) 3x(2) - 16x +5 = 0

    b) 2x(2) - 5x - 3 = 0

    c) 6x(2) - 13x + 6 = 0

    d) x(2) + 2x +4 = 0

    e) x(2) + 8x - 4 = 0

    f) x(2) - 4x +1 = 0

    the number in the brackets is the power (all squared). thank you for all the help!

    damon
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  2. #2
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    Quote Originally Posted by damonneedshelp View Post
    a) 3x(2) - 16x +5 = 0

    b) 2x(2) - 5x - 3 = 0
    You can factor, complete the square, or use the quadratic formula. Whichever you wish. I'll do these two in order. You can look up the quadratic formula on your own.

    Factoring:
    a) 3x^2 - 16x + 5 = 0
    There are many ways to factor. This one ALWAYS works if the quadratic can be factored at all.

    Multiply the coefficient of x^2 by the constant term:
    3 * 5 = 15
    Now find all the factors of 15:
    15 = (+/-)1*15
    15 = (+/-)3*5

    Which pair of these factors add up to the coefficient of the linear term? I get:
    15 = (-1)*(-15)
    -16 = -1 + -15

    Now split the linear term:

    3x^2 - 16x + 5 = = 3x^2 + (-1x - 15x) + 5 = 0

    Rearrange the parentheses:
    (3x^2 - x) + (-15x + 5) = 0

    Now factor from each set of parenthesis:
    x(3x - 1) + (-5)(3x - 1) = 0

    Note that both terms have a common 3x - 1. (This kind of thing will always happen.) So factor the common 3x - 1:
    (x - 5)(3x - 1) = 0

    Now set each factor to 0:
    x - 5 = 0 ==> x = 5

    3x - 1 = 0 ==> x = 1/3

    Completing the square.
    b) 2x^2 - 5x - 3 = 0

    Not all quadratics factor nicely. This method ALWAYS works.

    Rearrange the equation so that the constant term is on the RHS and divide through by the coefficient of the x^2 term:
    2x^2 - 5x = 3

    x^2 - (5/2)x = (3/2)

    Now, look at the identity: (x + a)^2 = x^2 + (2a)x + a^2. What we have in our equation is the first two terms of these. We want a value of a such that the LHS is a perfect square. So compare the coefficients of the linear terms:
    x^2 - (5/2)x
    x^2 + (2a)x + a^2

    Set 2a = -5/2

    a = -5/4

    So we wish to add a^2 to both sides of our equation:
    x^2 - (5/2)x + (-5/4)^2 = (3/2) + (-5/4)^2

    According to our identity the LHS is now in the form x^2 + (2a)x + a^2 = (x + a)^2:

    (x - 5/4)^2 = 3/2 + 25/16 = 49/16

    Take the square root of both sides:
    x - 5/4 = (+/-)sqrt(49/16) = (+/-)7/4

    x = 5/4 (+/-) 7/4

    So
    x = 3 or x = -1/2

    -Dan
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